Meta·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

Jun 2026

Summary

Meta software engineering interview with two coding problems back to back. Nothing too exotic but the tree BFS question had a follow-up on complexity that I fumbled a bit.

Questions Asked (2)

Given an array of positive integers and a target value, find the minimum length of a contiguous subarray whose sum is greater than or equal to the target. Return 0 if none exists. Expected O(n) time and O(1) space.

Algorithms & Data Structures

Author's notes

Sliding window clicked pretty fast for me since the array is all positive integers, which means shrinking from the left always makes sense.

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AI HintsAI Generated

Suggested Approach

Use a sliding window (two pointers) to maintain a window of elements whose sum is at least the target. Expand the right pointer to increase the sum, and once the sum meets or exceeds the target, shrink the window from the left to find the minimum length. Track the minimum length throughout and return it, or 0 if no valid window exists.

Pro tip: Clarify that the array contains only positive integers, which is crucial for the sliding window to work because it ensures the sum is monotonic as the window expands or shrinks. Also, mention that you'll handle edge cases like empty array or target larger than total sum.

1. Understand the problem and constraints

Restate the problem: find the minimum length of a contiguous subarray with sum >= target. Note that the array has positive integers, which allows the sliding window technique. Confirm expected time and space complexity.

2. Initialize pointers and variables

Set left pointer to 0, current sum to 0, and min length to infinity (or a large number). Iterate with a right pointer from 0 to n-1.

3. Expand and contract the window

Add the element at right to current sum. While current sum >= target, update min length with the current window size, then subtract the element at left from current sum and increment left.

4. Return the result

After the loop, if min length is still infinity, return 0; otherwise return min length.

Key Points to Mention

Sliding window technique with two pointers (left and right).
Time complexity O(n) because each element is visited at most twice.
Space complexity O(1) as only a few variables are used.
The importance of positive integers for the sliding window to work.
Handling edge cases: no valid subarray, empty array, target <= 0.
Updating the minimum length only when the sum condition is met.

AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.

Given the root of a binary tree, return its level-order traversal as a list of lists, where each inner list contains the node values at that depth from left to right. Implement it iteratively using BFS and walk through the time and space complexity.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

Classic BFS with a queue, nothing surprising in the implementation.

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AI HintsAI Generated

Suggested Approach

Start by clarifying the problem and edge cases, then describe the BFS approach using a queue and level-size tracking. Walk through a small example to illustrate, then implement the code iteratively, and finally analyze time and space complexity.

Pro tip: Mention that level-order traversal is the foundation for many tree problems (e.g., right-side view, zigzag traversal) and that BFS is preferred over DFS for level-based processing due to its natural level separation.

1. Clarify and Edge Cases

Confirm input/output format, discuss edge cases like empty tree, single node, and skewed tree. Ask if the tree is balanced or if there are constraints on node values.

2. Outline BFS Approach

Explain using a queue to process nodes level by level. At each iteration, record the current level size to process exactly that many nodes, collecting their values and enqueuing their children.

3. Walk Through Example

Trace the algorithm on a small binary tree (e.g., root with left and right children) to demonstrate how levels are formed and how the queue evolves.

4. Implement Code

Write clean, iterative code using a queue (e.g., collections.deque in Python). Ensure proper handling of null nodes and initialization of the result list.

5. Analyze Complexity

State that time complexity is O(N) since each node is visited once, and space complexity is O(M), where M is the maximum number of nodes at any level (worst-case O(N) for a full level).

Key Points to Mention

Use a queue (FIFO) to process nodes level by level.
Track the size of each level to separate levels in the output.
Time complexity: O(N) because each node is enqueued and dequeued exactly once.
Space complexity: O(M) where M is the maximum number of nodes at any level; worst-case O(N) for a full binary tree.
Edge cases: empty tree returns empty list; single node returns [[root.val]].
BFS is iterative and avoids recursion stack overhead, making it suitable for deep trees.

AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.